Advanced Factorization


Factorization is the decomposition of an expression into a product of its factors.

The following are common factorizations.

  1. For any positive integer \(n\), \[a^n-b^n = (a-b)(a^{n-1} + a^{n-2} b + \ldots + ab^{n-2} + b^{n-1} ).\] In particular, for \( n=2\), we have \( a^2-b^2=(a-b)(a+b)\).

  2. For n n an odd positive integer, an+bn=(a+b)(an1an2b+abn2+bn1). a^n+b^n = (a+b)(a^{n-1} - a^{n-2} b + \ldots - ab^{n-2} + b^{n-1} ).

  3. a2±2ab+b2=(a±b)2 a^2 \pm 2ab + b^2 = (a\pm b)^2

  4. x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzzx) x^3 + y^3 + z^3 - 3 xyz = (x+y+z) (x^2+y^2+z^2-xy-yz-zx)

  5. (ax+by)2+(aybx)2=(a2+b2)(x2+y2) (ax+by)^2 + (ay-bx)^2 = (a^2+b^2)(x^2+y^2). (axby)2(aybx)2=(a2b2)(x2y2) (ax-by)^2 - (ay-bx)^2 = (a^2-b^2)(x^2-y^2).

  6. x2y+y2z+z2x+x2z+y2x+z2y+2xyz=(x+y)(y+z)(z+x) x^2 y + y^2 z + z^2 x + x^2 z + y^2 x + z^2 y +2xyz= (x+y)(y+z)(z+x).

Factorization often transforms an expression into a form that is more easily manipulated algebraically, that has easily recognizable solutions, and that gives rise to clearly defined relationships.

Worked Examples

1. Find all ordered pairs of positive integer solutions (x,y) (x,y) such that 2x+1=y22^x+ 1 = y^2.

Solution: We have 2x=y21=(y1)(y+1)2^x = y^2-1 = (y-1)(y+1). Since the factors (y1)(y-1) and (y+1)(y+1) on the right hand side are integers whose product is a power of 2, both (y1)(y-1) and (y+1)(y+1) must be powers of 2. Furthermore, their difference is

(y+1)(y1)=2, (y+1)-(y-1)=2,

implying the factors must be y+1=4y+1 = 4 and y1=2y-1 = 2. This gives y=3 y=3, and thus x=3x=3. Therefore, (3,3)(3, 3) is the only solution.

2. Factorize the polynomial

f(a,b,c)=ab(a2b2)+bc(b2c2)+ca(c2a2).f(a, b, c) = ab(a^2-b^2) + bc(b^2-c^2) + ca(c^2-a^2).

Solution: Observe that if a=b a=b, then f(a,a,c)=0f(a, a, c) =0; if b=cb=c, then f(a,b,b)=0f(a, b, b)=0; and if c=a c=a, then f(c,b,c)=0 f(c,b,c)=0. By the Remainder-Factor Theorem, (ab),(bc), (a-b), (b-c), and (ca) (c-a) are factors of f(a,b,c) f(a,b,c). This allows us to factorize

f(a,b,c)=(ab)(bc)(ca)(a+b+c).f(a,b,c) = -(a-b)(b-c)(c-a)(a+b+c).

Note by Calvin Lin
6 years, 3 months ago

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For the worked example - 1 : there are 2 solutions : (3,3) and (3,-3)

(y+1)-(y-1) = 2

implies y + 1 = -2 or y + 1 = 4 and y -1 = -4 or y-1 = 2

Thus, the two solutions you have are (3,3) and (3,-3)

Kushashwa Shrimali - 6 years ago

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Thanks. I added in "positive integers".

Calvin Lin Staff - 5 years, 1 month ago

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Thanks for this.

Anuj Shikarkhane - 6 years ago

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I did not understand why should (y+1) - (y-1) = 2 ?
Can anyone explain ?

Raven Herd - 5 years, 1 month ago

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What do you not understand?

What do you think (y+1)(y1) (y+1) - (y-1) is equal to ?

Calvin Lin Staff - 5 years, 1 month ago

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I did not see it correctly . My fault ! Sir, I have a problem. I want to learn Number theory as is organized here on Brilliant but I don't follow anything beginning from modular inverses.I have tried the wikis but I still don't follow .Please suggest something.

Raven Herd - 5 years, 1 month ago

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How is their (y+1)-(y-1) difference 2?

Ishan Saxena - 6 years, 1 month ago

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(y+1)(y1)(y+1) - (y-1) =y+1y+1= y +1 - y + 1 =1+1=2= 1+1= \boxed{2}

Ankita Mehra - 6 years, 1 month ago

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(Y+1)-(y-1) = y+1-y+1= 2

Abhishek Gupta - 6 years, 1 month ago

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sandeep patel - 6 years, 1 month ago

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How to think that both difference is 2. What is the main moto behind thinking such that????

PRASHANT KOCHAR - 6 years ago

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